Transcript Ch 3 Single Particle Motion - University of Massachusetts

Lecture_5_Single_Particle_Motion_1
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Ch 3 Single Particle Motion
Consider the Lorentz force when E  x, t  and B  x, t  are specified.
 Is this normally the case?? 
dv
 q E  v  B
dt
dx
 v  x, t 
dt
To determine the motion of a single charged particle in the fields
m
we can solve above DEs. Consider different situations:
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3.2 Uniform magnetic field, and E = 0 :
dv
m
 qv  B
dt
It is customary (and very useful) to set v  v //  v  (natural comp.)
Note that  v //  v    B  v   B. Then
dv //
dv 
m
 0, m
 qv   B or
dt
dt
dv  q
 v   B  v   b with
 3.6 
dt
m
qB
B

, b
 is the gyrofrequency (Lamor frequency)
m
B
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Orient the z axis of the cartesian coordinate system in the b direction.
Then
x y z
v   v x  v y , v  v z , b  z, and v   b  vx v y vz
0 0 
dv y
dvx
dvz
 v y ,
 vx ,
0
dt
dt
dt
\These are coupled DE's that can be "uncoupled" by differentiating:
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dv y
dvx
 v y ,
 vx . Differentiate re t:
dt
dt
dv y
d 2v y
d 2 vx
dvx

,
 
. Substitute:
2
2
dt
dt
dt
dt
d 2v y
d 2 vx
   vx  ,
   v y 
2
2
dt
dt
Solve ordinary DE
 3.7  8 
d 2 vx
2


vx  0. Try
2
dt
vx  v0 exp i   t    or the Re thereof.
d 2 vx
2
2
2

i

v
ex
p
i

t




v
exp
i

t




vx . OK






0
0
2
dt
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From  3.7  :
1 d 2 vx
1
v y   2 dt    2 vx dt 
 dt

  vx dt  ivx
v y  iv0 exp i   t    . The minus sign for the electron.
Take the real parts:
vx  v0 cos   t   
 3.12a 
 3.12b 
v y  v0 sin   t   
v 2  vx2  v y2  v02 .
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vx  v0 cos   t    ,
v y  v0 sin   t    . Integrate:
v0
v0
x  x0 
sin   t    , y  y0 
cos   t   


This is a cicular motion in the x,y plane.
Discuss right/left hand circles.
dvz
We had for the z component
 0. Therefore in the
dt
z direction, the charge moves with constant velocity v // :
dz
d 2z
z  z0  v// t 
 v//  2  0
dt
dt
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v
v
x  x0 
sin   t    , y  y0 
cos   t   


Lamor or gyro radius:
rL 
 x  x0    y  y0 
2
2
v


v mv
rL 

 qB
 3.15
The circumference of the gyro orbit is 2 rL , and the time for 1 orbit:
2 rL 2
m
T

 2
v

qB
 3.14 
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Pitch angle and magnetic moment
The perp velocity v   vx2  v y2  v0 is constant, and so is v // , so the
v v0
the ratio is constant: tan  

,
v// v// 0
 is called the pitch angle.
The magnetic moment of a current loop is
 M  I A where I=current, A=area.
q
1
For gyrating charge q, the current is I  
q 
T 2
 v
The area is A   r   

2
L



2
2
2
1 q v 1 q mv
M  I A 

,
2 
2 qB
1 2
mv
W
M  2
 
B
B
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 3.17 
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3.3 Uniform External Force
We now assume uniform B field and uniform external force F.
dv
m
 q  v  B  F
dt
We set F  F//  F . Then
dv// t
m
 q  v  B //  F//  F// ; since  v  B //  0
dt
F//
v// t  v// 0  t  v// 0  a// t
m
dv t
m
 q  v t  B   F  q  v t  B   F
dt
To solve  3.21 , we assume that we can express
v t  t   v c  t   v F
 3.20 
 3.21
 3.22 
where v c  t  describes the cyclotron motion, and v F the drift
motion due to the external force.
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dv c  t 
We know from (3.6) on p.3 that
 vc  Ω
dt
dv F
Also
 0 if we assume that v F is constant.
dt
Then  3.21 becomes:
d  vc  v F  q
1
  v c  v F   B   F or
dt
m
m
=W
q  1

v c  Ω   v c  v F   B   F
m  m

1
v c  Ω  v c  Ω  v F  Ω  F
m
1
 0  v F  Ω  F  Solve for v F by applying Ω 
m
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Ω  v F  Ω  v F  Ω  Ω   Ω  v F  Ω   v F 2
since v F  Ω  0  v F  Ω, see p.10, 3.22 
1
1
 v F    Ω  F   Ω  F, or
m
m
2
FΩ FB
B qB B
vF 

recall Ω   
2
2
m
qB
B m B
 3.24 
For time independent F and B, v F is constant in the
direction of F  B if q is positive, otherwise in
opposite direction.
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If F+ is the force on the ions, and F- the force on the electron, then
F  B
F  B
v Fi 
, v Fe 
and the current density becomes:
2
2
eB
eB
F B F B
J F  n  qi v Fi  qe v Fe   en  v Fi  v Fe     2   2 
B 
 B
  F  F   B 
JF  n 
 3.25

2
B


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3.4 Uniform Electric and Magnetic Fields
The electric force is F  qE. Then from (3.24):
FB EB
vE 

2
qB
B2
 3.26 
The "E  B " drift velocity is independent of charge,
i.e., same for i and e (no current!). Recall the total perp velocity is
v  vc  v E
Note that E  B  E  B.
When the positive ion moves in the direction of E , it is accelerated,
v increases, and rL  v  increases. rL becomes smaller when the
ion moves opposite to E , etc. This causes a drift  to B (Fig.3.3).
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